Question

Suppose that 14 bluebirds and 11 robins are trying out for a choir that will have 7 members.How many ways can this choir be formed if it must have more bluebirds than robins?Note:no two animals are exactly alike.

Answer 1

Answer:311740

Step-by-step explanation:

Given

There are 14 blue birds and 11 robins

We have to choose 7 members such that there are more blue birds than robins

This can be possible in 4 ways i.e.

7 BB,6 BB and 1 R, 5 BB 2 R,4 BB 3 R

Therefore total no of ways are

$=^{14}C_7+^{14}C_6\times ^{11}C_1+^{14}C_6\times ^{11}C_2+^{14}C_4\times ^{11}C_3$

$=311740\ ways$

Answer 2

Answer:

Step-by-step explanation:

number of bluebirds = 14

number of robins = 11

Total selected birds = 7

There are several ways to select the birds

(1) Selection of 7 blue birds out of 14 = $^{14}C_{7}=\frac{14!}{7!\times 7!}=6864$

(2) selection of 6 blue birds and 1 robin = $^{14}C_{6}\times ^{11}C_{1}=\frac{14!}{6!\times 8!}\times \frac{11!}{1!\times 10!}=33033$

(3) selection of 5 blue birds and 2 robins = $^{14}C_{5}\times ^{11}C_{2}=\frac{14!}{5!\times 9!}\times \frac{11!}{2!\times 9!}=110110$

(4) selection of 4 blue birds and 3 robins = $^{14}C_{4}\times ^{11}C_{3}=\frac{14!}{4!\times 10!}\times \frac{11!}{3!\times 8!}= 165165$

So, the total number of selections to have more blue birds than robins =

= 6864 + 330330 + 110110 + 165165 = 612469